91Ó°ÊÓ

The revenue function describes how much income a company can generate from selling its products. In our exercise, the price for each chair is determined by the quantity produced, which is given by the price function: \( p(x) = 200 - 0.15x \).
To find the total revenue, we multiply the price per chair by the number of chairs sold, resulting in the revenue function: \( R(x) = (200 - 0.15x) \times x = 200x - 0.15x^2 \).
This function helps visualize how changes in production levels can affect the company's income.

• As production increases, revenue initially rises since more chairs are being sold.
• Increased production beyond a certain point could decrease revenue per unit due to the price reduction per chair.

Understanding the revenue function is crucial for analyzing any firm's financial health, as it provides insights into how production volume impacts earnings.

The cost function represents the total cost associated with producing a specific number of items. This includes both fixed and variable costs.
In the exercise, the cost involved in producing \(x\) chairs changes when the additional machine is introduced:

• For up to 500 chairs, the cost function is \( C(x) = 5000 + 6x - 0.002x^2 \).
• With the additional machine (500 to 750 chairs), the cost function becomes \( C(x) = 9000 + 6x - 0.002x^2 \).

Fixed costs are represented by the constant term in the function, whereas coefficients of \(x\) and \(x^2\) represent variable costs. The quadratic term explains the increasing cost per additional chair, further explaining the decrease in economic efficiency as production scales up.
Aligning production with cost efficiency is vital for profit maximization and ultimately for deciding on the new machine.

Derivative analysis involves determining how a function changes as its input changes. For profit maximization, it's important to assess the rate of change of the profit function.
The profit function is the difference between revenue and cost, \( P(x) = R(x) - C(x) \),
from which the derivative is taken: \( P'(x) \).
The derivative \( P'(x) = -0.296x + 194 \) indicates where profit increases or decreases
or if it remains stable.

• A positive derivative means profit increases with more production.
• A negative derivative signals a decrease in profit with more production.
• Setting the derivative to zero finds the production level yielding maximum profit.

Understanding this concept helps companies strategically adjust production to operate efficiently at peak profitability.

An optimization problem involves finding the best solution from a set of feasible solutions. In this scenario, the goal is to determine whether the company should invest in the additional machine and what production level will maximize their profit.
Using derivative analysis, the optimal production level is found when
\( P'(x) = 0 \) to determine the critical points. For both scenarios (before and after the machine purchase), this occurs at
\( x \approx 655 \).
Evaluating these findings means comparing potential profits:

• Calculate the profit at current capacity with
  \( x = 500 \).
• Assess the new machine's profitability by evaluating production at
  \( x = 655 \).

In essence, this optimization ensures a decision that maximizes profits while aligning with the company's capacity and investment strategy. The critical analysis of both scenarios leads to an informed decision about machine acquisition.